求函数y=(x^2-3x-10)/(x^2+3x+2)的值域,而y=[(x^2+3x+2)-6x-12]/(x^2+3x+2)=1-6*[(x+2)/(x^2+3x+2)]=1-6*[(x+2)/(x+2)(x+1)],x≠-1,-2则y=1-[6/(x+1)]故y≠1,当x=-2,y=7,而x≠-2,故y≠7∴其值域为{y|y≠-2,y≠7}...